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76 lines
5.5 KiB
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76 lines
5.5 KiB
Markdown
---
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title: "HW2"
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aliases:
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tags:
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- math202
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- assignment
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---
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# Assignment 2
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Jet Hughes 9474308
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# 1.
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Let $V = P_2(\mathbb{R})$ with the ususal vector addition and scalar multiplication. For each of the following subsets of $V$, either prove that it is a basis of $V$ or explain why it is not a basis of $V$ . You may use any result from class.
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(a) $\{759, 20+2x+43x^2\}$
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(b) $\{1-x, 2x^{2},3+x^2\}$
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(c) $\{2x, 4+2x-x^{2}, -4-6x+x^2\}$
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(d) $\{-1+3x, 1+x^{2,}x-3x^{2,}4+4x-11x^2\}$
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Since $P_2(\mathbb{R})$ has dimension 3, by default all bases of $P_2(\mathbb{R})$ have three elements. Hence (a) and (d) cannot possibly be bases of $P_2(\mathbb{R})$
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(b) Since this set has 3 vectors and $P_2(\mathbb{R})$ has dimension 3, it is enough to check either that is is linearly independent or that it spans $P_2(\mathbb{R})$.
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To show linear independence, if $a(1-x)+b(2x^2)+c(3+x^2)=(0x^2 + 0x + 0)$, we have $2b+c=0$, $-a=0$ and $a+3c = 0$. So $a=0$ which implies $c=0$ which then implies $b=0$. So the only linear combination equal to the zero vector is the one where $a=b=c=0$, hence this set in linearly independent. Since it is linearly independent and has three vectors, its span is a subspace of $P_2(\mathbb{R})$ of dimension 3, i.e., all of $P_2(\mathbb{R})$
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(c) Since this set has 3 vectors and $P_2(\mathbb{R})$ has dimension 3, it is enough to check either that is is linearly independent or that it spans $P_2(\mathbb{R})$.
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To show linear independence, if $a(0 +2x+0x^2)+b(4+2x-x^2)+c(-4-6x+x^2)=(0+0x+0x^2)$, we have $4b-4c=0$, $2a+2b-6c=0$ and $-b+c = 0$. So $b=c$ which implies $-c+c=0$ and $2a+2c-6c=0$ so $a=2c=2b$. But this does no force $a,b,c$ t obe zero; we could have e.g., $a=2$ and $b=c=1$. Therefore we have a linear combination of the vectors that gives the zero vector when the coefficients are not all zero. So it is linearly dependent, and therefore can't be a basis.
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# 2.
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Let $V$ be a vector space such that $dim(V)= 10$. Let $U$ and $W$ denote subspaces of $V$.
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(a). What is the max value of $dim(U\cap W)$, assuming $dim(U)=7$ and $dim(W)=4$ ?
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The largest possible dimension of $(U\cap W)$ is 4 which occurs when $W$ is entirely contained in $U$.
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(b). What is the min value of $dim(U+W)$
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$dim(U\cap W) \leq 4$ so from the formula $dim(U+W) = dim(U) + dim(W) - dim(U \cap W)$ we have $dim(U+W) \geq 7 + 4 - 4= 7$.
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(c). Max value of $dim(U+W)$
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We have $dim(U+W) \leq dim(U) + dim(W) = 11$. But since $U+W$ is a subspace of $V$ which has dimension $10$ , we must have $dim(U+W) \leq 10$. To show that this can be achieved, we could let $\{v_1,...,v_{10}\}$ be a basis for V, and choose $U = span\{v_1, v_2, v_3, v_4, v_5, v_5, v_7)\}$ and $W=span\{v_7,v_8,v_9,v_{10}\}$. Then $U+W$ will contain the basis for $V$ and hence $U+W=V$. So the maximum possible value of dim(U+W) is 10.
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(d). Since $dim(U+W) \leq 10$ and $dim(U+W) = 11 - dim(U\cap W)$, we have $dim(U\cap W) \geq 11-10= 1$
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(c). No, because if $U+W$ is direct then $dim(U+W) = 0$, but we showed that $dim(U\cap W) \geq 1$
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# 3.
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Let $U$ = $\{p\in P_2(\mathbb{R}): p(x)$ is divisible by $x-3\}$. Then U is a subspace of P2
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(a) Find a basis of U.
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Suppose a polynomial of degree 2 is divisible by $x-3$ . It can be written as $(x-3)(ax+b)$ where $a,b \in \mathbb{R}$. Then we have a basis $\{(x-3), x(x-3)\}$
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To prove this is a basis we need to show that it is linearly independent and that is spans $U$ . To show linear independece if $a(x-3)+b(x^2+3x) = (0x^2 + 0x + 0)$, we have $-3a=0, a-3b=0$ and $b =0$. So $a=b=0$ and the only linear combination equal to the zero vector is the one where $a=b=0$, hence this set is linearly independent. Since U must have dimension 2, it must be a spanning set and therefore must be a basis.
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(b). Find another subspace $W$ of $P_2(\mathbb{R})$ such that $U+W$ is a direct sum.
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Need to find $W$ where $U\cap W = \{0_v\}$ and $U+W = P_2(\mathbb{R})$. We have a basis of $U = \{(x-3), (x^2-3x)\}$. We can adjoin a basis $\{1, x, x^2\}$ of $P_2(\mathbb{R})$ to get a spanning set $\{x-3, x^2-3x,1, x, x^2\}$ of $P_2(\mathbb{R})$.
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The 3rd vector is not a linear combination of the first two. The 4th vector is a linar combination of the 1st and 3rd vectors: $(x)=(x-3)+3(1)$. So we eliminate the 4th vector. The 5th vector is a linear combination of the 2nd 3rd and 1st vector: $(x^2) = (x^2-3x) + 3(x-3) +9(1) = x^2 - 3x +3x -9 + 9 = x^2$. So we eliminate the 5th vector. This leaves the basis $\{x-3. x^2-3x,1\}$ of $P_2(\mathbb{R})$. Let $W=span\{(1)\}$. To show that the sum $U+W$ is direct we compute $U\cap W$. $W$ consists of all polynomials of degree 0, and $U$ consists polynomials divisible by $x-3$. The only polynomial of degree 0 divisible by $x-3$ is $0$. So $U\cap W = \{0\}$, hence the sum is direct.
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To show the sum is equal to $P_2(\mathbb{R})$, we use the formula $dim(U+W) = dim(U) + dim(W) - dim(U \cap W)=2+1-0=3$ since $U+W$ is a subspace of $P_2(\mathbb{R})$ of dimension 3, we must have $U+W=P_2(\mathbb{R})$.
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# 4.
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Let $V$ and $W$ be vector spaces over $\mathbb{R}$ and $T: V → W$ a linear transformation. Let $\{v_1,...,.v_n\}$ be a basis for $V$
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(a) Since T is linear and $\{v_1,...,.v_n\}$ is s basis of $V$ $range(T) = T(V) = \{T(v): v \in V\}$. So $V= a_{1}v_{1}+...+a_{n}v_{n}$ for $a_1,...,a_n \in \mathbb{R}$. Then $T(V) = T(a_{1}v_{1}+...+a_{n}v_{n}) = a_{1}T(v_1)+...+a_{n}T(v_n)$ So $range(T) = T(V)$ is a linear combination of ${T(v_1),...,T(v_n)}$. So this must be a spanning set of $range(T)$
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(b) let $V = \mathbb{R}^2$ and $W = \mathbb{R}^2$. Let $V$ have a basis $\{(1,0), (0,1)\}$ Define $T:V→W$ by $T(x,y) = (x,y+3)$
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# 5.
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(a)
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(b) |